Free arrangements and relation spaces
Discrete Comput Geom
Free Arrangements 0 1
Relation Spaces 0 1
0 2 Department of Mathematics, University of WisconsinMadison , Madison, WI 53706 , USA
1 1Department of Mathematics, University of Kansas , Lawrence, KS 66045 , USA
Yuzvinsky [7] has shown that free arrangements are formal. In this note we define a more general class of arrangements which we call kformal, and we show that free arrangements are kformal. We close with an example which distinguishes kformal arrangements from formal arrangements. Let be a field a n d let V be a n /  d i m e n s i o n a l v e c t o r s p a c e o v e r K. A hyperplane in V is a c o d i m e n s i o n 1 s u b s p a c e o f V. A n arrangement ~t in V is a finite set of h y p e r p l a n e s . L e t {xl . . . . . xt} be a basis for t h e d u a l V* a n d let S be the s y m m e t r i c a l g e b r a of V* w h i c h is i s o m o r p h i c t o the p o l y n o m i a l a l g e b r a KIxl . . . . . x J . T h e n e a c h h y p e r p l a n e H in I / h a s a defining f o r m

1. Introduction
orR = a l x I 'k "'" k a/X/
with ker(gn) = H, u n i q u e u p to a c o n s t a n t multiple. T h u s a n a r r a n g e m e n t d c a n
* The first author was supported in part by a U.S. Department of Education Fellowship. The
second author was supported in part by the National Science Foundation.
be described by the product of such forms. Given an arrangement d , let
Q = Q ( ~ ) =
II ct..
H E ~r
Note that Q is unique up to constant multiple. The arrangement d is completely
determined by Q.
Let D(~r be the Smodule which consists of all derivations 0: S * S such that
0(Q) is a multiple of Q. We call D(~r the module of ~r When D(~r
is a free Smodule, d is said to be free. There are some interesting connections
between the lattice L ( d ) of intersections of elements of d and the Smodule D(d).
For a survey, see the book by Orlik and Terao [
3
]. In this work we study these
connections further. In particular, we are interested in the study of formal
arrangements, introduced by Falk and Randell [
1
]. An arrangement d is formal
if all linear dependencies among the defining forms of the hyperplanes of d are
generated by dependencies corresponding to the codimension 2 subspaces in L(d).
Using techniques developed in [
6
], Yuzvinsky has shown in 17] that free
arrangements are formal. In this paper we introduce a more general class of arrangements
called kformal arrangements. By definition, d is 2formal if d is formal. The
main result of this paper is that if d is free, then d is kformal for 2 < k < r  1,
where r is the codimension of the common intersection of all the hyperlanes of
d . Our proof does not use techniques from [
6
]. This theorem imposes strong
conditions on the ways in which the hyperplanes of a free arrangement intersect
each other. In particular, we have some combinatorial inequalities for free
arrangements. We close the paper with an example of an arrangement which is
2formal (formal) but not 3formal. This distinguishes kformal arrangements from
formal arrangements.
2. Preliminaries
We begin with some definitions and a few basic facts which are used throughout
this paper. For further background, see [3]. Fix an arrangement d . Let n = I~r
Let S and Q ( d ) be as described in the introduction. We call Q ( d ) the defining
polynomial of ~r
Definition 2.1. Let L = L(~r be the set of intersections of elements of ~ . Define
a partial order on L by
X ~ Y
~
Y ~ X .
Note that this is reverse inclusion. Thus V is the unique minimal element.
Definition 2.2. Define a rank function on L by r(X) = codim X. Thus r(V) = 0
and r(H) = 1 for H 9 ~ . Call H an atom of L. Let X, Y 9 L. Then their meet X ^ Y
is equal to ('] { Z 9 u Y ~ Z}. Their join X v Yis equal to X n Y.
[]
Proof.
See Proposition 4.8 of [3].
Corollary 2.8. I f g~ ~ ~ , then D ( ~ ) ~_ D(~).
Thus if X, Y 9 L ( ~ ) , with X __ Y, then D(~r) _ D(~x).
Definition 2.9. An arrangement ~ is called a free arrangement if D ( ~ ) is a free
module over S.
Theorem 2.10. I f d is free, then ~ x is free for each X 9 L.
3. Formal Arrangements
Fix an arrangement ~. Choose a defining form aH for each H ~ ~ . We wish to
study the various relations among these forms. We consider the vector space E(~/)
which has a basis consisting of the symbols {eH[H 6 ~/}. We have a map
= 9
H e d
Ke.v*
given by eu ~ uu Now define F ( ~ ) to be the kernel of this map. We call F(~/)
the relation space and we refer to elements of F ( d ) as relations. Then we have an
exact sequence
Definition 3.1. An element w = ~ crier of E(~/) has length p if precisely p of the
ca 6 K are nonzero. We write p = length(w).
Definition 3.2. Let n = I ~ 1 . Let 2 _ p _< n  1. We define Fp(~) to be the
subspace of F(,~/) spanned by all relations of length no greater than p + 1:
Fp(~l) = <{w = ~ cHeB ~ F(~/)llength(w) _< p + 1}).
Definition 3.3.
2generated.
We say ~ is poenerated if F ( d ) = Fp(~/). Call ~ / f o r m a l if d is
Remark 3.4. Let r = r(~). For r < p _< n  1, Fp(~/) = F(~). Thus all
arrangements are rgenerated.
Note that M formal is the strongest of the above conditions. Note that if r < 2,
then it is always the case that F(M) = F2(~/). Thus, for the rest of the paper, we
assume that r > 3.
Remark 3.5. F o r each X ~ L, we have an inclusion F2(,~/x) ~ F(~). If ~ / i s formal,
then each relation in F ( ~ ) can be written as a sum of relations in F2(,.Q/). Thus
, d is formal if and only if we have a surjection
~2: ( ~
X e L
r(X): 2
F2(,~x) '~ F(~/) ' O.
Note that if r(X) = 2, then F 2 ( d x ) = F(..q/x).
Free Arrangementsand RelationSpaces 53
Example 3.6. Consider the real 3arrangements ~r and ~r defined by
Q1 = X 1 X 2 X 3 ( X 1  X2)(X2  X3)(X1  2X3)
and
Q2 = X 1 X 2 X 3 ( X I "~ X2 dl X3),
respectively. Then ~r is formal, whereas ~r is not.
Remark 3.7. Although each linear dependency among the defining forms of the
hyperplanes of ~r is associated to a particular subspace in L, it is not true that
the property of being formal depends only on L (see Example 2.2 of [7"1). Note,
however, that ~r formal does impose some conditions on L. By Remark 3.5, we
have
r, ( l ~ x l  2) _> n  r ( ~ ' ) .
XcL
r~X) = 2
In [
1
] Falk and Randell asked if free arrangements are formal. This question
was resolved by Yuzvinsky:
Theorem3.8 [7, Corollary 2.5]. I f ~t is free, then ~ is formal.
We reprove this by proving our main theorem in the next section.
Using this theorem, we can give some easy lattice conditions which are
necessary for an arrangement to be free. In addition to the inequality in Remark
3.7, we have the following proposition. The proof is straightforward and is left to
the reader. Call H ~ ~r a separator if r(~r  {H}) < r(~r
Proposition 3.9. Let ~ be an arrangement. Fix H ~ d and let ~ ' = ~  {H}.
(i) I f H is a separator, then ~ is formal if and only if ~t' is formal.
(ii) Suppose H is not a separator, l f ~ / is formal, then H contains a rank 2 element
of H ~ ' ) .
Example 3.10. The real 4arrangement ~r defined by
~2 = xlx2x3x4(Xx
 x 2 ) ( x l + x3  x4)
is not free.
Proof. The plane H = ker(xl + xa  x4) is not a separator, and does not contain
any rank 2 subspace of L ( ~ r {H}). Thus by the previous proposition and
Theorem 3.8, ~r is not free. Note, however, that H does contain the rank 3 subspace
ker(xl) n ker(xa) c~ker(x4). []
4. A Generalization of Formal Arrangements
We now define the class of kformal arrangements and give a generalization
of Theorem 3.8.
Suppose ~r is formal. Then we have an exact sequence
( ~
XeL
r(X) = 2
0 ~ R 3 ( ~ ) ~
F2(~x) ~ F(~) ~ O,
where Ra(0ff) is the kernel of the map n2. Thus Ra(~r ) is a space of relations
among the relations corresponding to the rank 2 elements of L. Note that,
for any X e L, we have an inclusion Ra(dx) ~ R3(~r ). If the map
( ~
XEL
r(X) = 3
Rs(~/x) ~ Ra(~t )
is surjective, then we call ~ 3formal. We can continue in this fashion to
define kformal for k > 3. We make this idea precise in the following definitions.
Definition 4.1. Let R o ( ~ ) = T(~t)*.
Remark 4.2.
Note that, for each X E L, there is a restriction map
Ro(~x) ~ Ro(~r
Definition 4.3. For 1 _< k _< r, define Rk(d~ ) recursively to be the kernel of
the map
X~L
r(X) = k  1
Rk  l(.~x) ~ R k_ 1(.~),
where the maps nk are defined as follows. For each k _> 1 and for each Y e L
with r(Y)_> k, we have a map ik(Y): Rk(~r)~Rk(a~,), which can be viewed
as inclusion. The map 7~k is a sum of these maps. Also, we have a commutative
diagram:
0
0
' Rk+l(~r)
' ( ~ Rk((~r)x) ~kt~ Rk(~y )
X<Y
' Rk+ l(a~r
'
Rk(~x)
,,(d) Rk(~).
Since X < Y, we have (Mr)x = Mx. The middle vertical map Jk(Y) is a direct
sum of identity maps and zero maps, so it can be viewed as inclusion.
N o t e that if r ( X ) = 0, then we have X = V. Thus R I ( M ) is the kernel of
the restriction m a p V * ~ T(M)*, so R I ( M ) = T(M) ~ N o t e also that we have
identifications R dMn) = T(Mn) ~ ~ H ~ ~ ~ e n. Hence R2(M ) can be identified with
the kernel of the m a p
~ e n ~ T(M) ~
so we see that R2(M ) ~_ F(M).
Definition 4.4. F o r 2 < k < r  1, define the class of kformal arrangements as
follows:
(i) An arrangement d is 2formal if M is formal.
(ii) F o r k > 3, M is kformal if d is (k  1)formal and the m a p
XeL
r(X) = k
is a surjection.
O u r main theorem states that if d is free, then M is kformal for 2 < k < r  1.
To prove this theorem, we first relate the vector spaces Rk(M) to the Smodule of
derivations D(M). We define Smodules Dk(M ) for 0 < k < r. O u r construction is
analogous to that of the vector spaces Rk(M ).
Definition 4.5. Let D o ( d ) = D(M).
Recall that for each X e L, we have an inclusion m a p Do(M ) , Do(Mx). D e n o t e
this m a p by tpo(X).
Definition 4.6. F o r 1 ~ k < r, define the Smodules Kk(d~[) and Dk(M) recursively
to be the kernel and cokernel respectively o f the m a p
rk, = Tk I(M): Dk I(M) ~
Dk t(Mx),
@
XeL
r(X) =k  1
where the m a p Zk is a sum of maps ~k(Y): Dk(~C)*Dk(Slr). F o r Y 6 L
with
r(Y) > k  1, these m a p s ~pk(Y) are defined by the following diagram:
( ~
Xr
v(X) = k  1
~ )
X<Y
r(X) = k  1
Dk_,(sr
,k,(~)
Dk_,(Mx )
, Dk(M )
I tPItI(Y)
Pit I(Y)
r
D k 1(~r ~*'(~'~
Dk,((~Cr)x)
, Dk(Mr)
, 0
, 0 .
R e m a r k 4.7. Note that Kt(~qr = 0 for any arrangement, since it is the kernel of
the inclusion m a p Zo: Do(M) + Derx(S). Also K 2 ( , ~ ) = 0, for if [0] is in the kernel
of the m a p
z,:Dl(~r
0
H~,rr
D~('ClH),
then 0 e Do(~Cn) for each H e ~ .
R e m a r k 4.8. If ~r is the empty arrangement, then D l ( d ) = 0, and hence
Ok(,.r16)2 = 0 for 1 ~ k _< r.
Remark 4.9. Note that D1(~r ) = Derx(S)/D(~).
g: T ( ~ )  , Do(~')o given by
We have an isomorphism
l
g(v) = ~
/=1
x,(v)D,
for each v e T(~r
Thus DerK(S)o ~ V and D1(~r
"~ V / T ( ~ ) .
Proposition 4.10. For 0 < k < r, we have Dk(d)* ~Rk!~).
Proof. Recall that Ro(~r ) = T(~r By the r e m a r k above, we have an
isomorphism fo(~r = g*: Do(d)* * Ro(~t) for any arrangement ~ . For k ___O, consider
the following diagram:
0
, R,+,(z~r
"it, Rk(.~).
r(X) = k ~
XEL
,
First note that the rows are exact. It is straightforward to show that this
diagram commutes for k = 0, and hence we can construct an isomorphism
f1(~r D I ( ~ ) * ~ R1(~r ), for any arrangement ~r We continue via induction,
noting that it is sufficient to show that the following diagram commutes for each
X e L with r(X) = k:
Dk(dX), ~Cx~ Dk(d)~
I fk(~eIX)
I fk(~
Rk(~/x)
,,(x) Rk(d).
By chasing this diagram, we can show that, for each k < r, the diagram commutes
and hence we have an isomorphism fk(~r Dk(~r ~ Rk(~t ). The idea is to combine
the above diagram with those from Definitions 4.3 and 4.6. []
We may view the correspondence X ~  ~ D ( X ) = D(~x) for X e L as a
contravariant functor on L, because there is the inclusion m a p D(X) , D(Y) whenever
X > Y. Note that D(T(~r D(~r and D ( V ) = DerK(S). It turns out that this
functor is local (in the sense of Definition 6.4 of [
4
]):
Proposition 4.11. Let X e L and ga ~ Spec(S), and define
N H.
He~tx
~ue~
Then the localization at ga of the inclusion map D(X) ~ D(X(ga)) is an isomorphism.
Proof. This is contained in Example 4.123 of [3].
[]
L e m m a 4.12. For 0 <_k <_r, the functors DR(X) = Dk(dX) for X ~ L are local.
Proof. This follows using induction and the fact that localization preserves exact
sequences. []
The following result is essentially L e m m a (5.15) of 15], but since our setup is
a bit different, we give a detailed p r o o f here.
Proposition 4.13. I f s# is free, then, for 1 <~k < r, we have
(i) pds D k  l ( d ) = k  1,
(ii) ht go = k  1 for each fJ e Asss Dk l ( d ) ,
(iii) Kk(~r = O.
Before the proof, let us recall a fundamental fact from commutative algebra.
F o r details (in the local ring case) see P r o p o s i t i o n 3.9 in C h a p t e r VI of [2].
Theorem 4.14. For any finitely oenerated oraded Smodule M,
depth s M <
m i n { l  h t p } <
$~9 Asss M
m a x { l  h t g o } =
~oe Asss M
max { l  h t g o }
~ e Supps M
= dim s M.
Here, depth s M is defined with respect to the maximal h o m o g e n e o u s ideal of S.
P r o o f o f 4.13. We use induction. W h e n k = 1 we k n o w pds Do(M) = 0 since M
is free, so (i) is clear. Also, since Do(M ) is free, the annihilator of each 0 ~ Do(M )
is the zero ideal. Thus Asss Do(M ) consists of the zero ideal, and now (ii) follows.
In R e m a r k 4.7 we saw that K I ( M ) = 0 so (iii) holds.
Suppose the result holds for some k (1 < k < r  1). Then, since K k ( M ) = 0, we
have an exact sequence
XEL
r(X) = k 1
Dk_ l(Mx) * Dk(M ) *O.
T h u s pds Dk(.~[) <k. N o w let ~o e Spec(S) with ht ~ _< k  1. We write M~ =
Mr(~,Xv ). T h e n we have the exact sequence
o ,
x(M )o
0
XeL
r(X) = k  1
Dk_ l(Mx{r))~ ~ Dk(M)to ~ O.
N o t e that there is at m o s t one X e L of rank k  1 (in the case when ht go = k  1)
with O k_ l(Jdx(~)) :f: 0. Thus either
Dk I(M~)~ ~
Dk l(MX{r))~
( ~
XEL
r(X) =k  1
is an i s o m o r p h i s m o r D k_ l(MX{~)) = 0 for all X e L with r(X) = k  1. In either
case, Dk(M)~ = 0. This shows that, for a n y go e Supps Ok(,~f), we have ht ~ >_ k,
and hence, by T h e o r e m 4.14, we k n o w depth s Dk(M ) _< dim s Dk(M ) <_ I  k.
NOW by the A u s l a n d e r  B u c h s b a u m formula (for graded modules), we have
l = pd s D~(M) + depth s Dk(M) <pds Dk(M) + l  k.
We have seen that pd s Dk(M ) <_ k, so it follows that pds Dk(M) = k. This shows
(i). Also,
l  k = depths D k ( d ) ~ dims Dk(M) <_ l  k.
Thus Dk(d~r is CohenMacaulay, and, by Theorem 4.14, each go e Asss Dk(,J~) has
height k. This shows (ii).
To show (iii), consider the sequence
Dk(~X).
Suppose there is some ~o e Asss Kk+ 1(~r Then, since Asss Kk+ 1 ( ~ ) Asss Dk(,~r
we know that ht 9 = k by (ii). We localize to obtain the exact sequence
0 * Kk+ 1(~)~ ~ Dk(~'~)~ ~
D~(.~xt~))~.
( ~
XeL
r(X) = k
Since ht go = k, there is at most one X e L with r(X) = k such that Dk(~X(~o))# 0,
so either the m a p
D,(~r
Dk(~c4X(,o))~,
~
XeL
r(X) = k
is an isomorphism o r Dk(dp) = 0. This shows that Kk+ l ( ~ r = 0. This is
impossible, however, since Asss KkÂ§ 1(~r Supps Kk+ l ( d ) [2, Proposition 2.9 in
Chapter VI]. Thus it must be the case that A s S s K k + l ( d ) = ~ . This implies that
K,+ 1(~r = 0 [2, Proposition 2.4 in Chapter VII. Hence (iii) is established and
the proof is complete. []
Now we have our main theorem:
Theorem 4.15. I f d is free, then d is kformal for 2 < k < r  1.
Proof. Suppose 2 < k < r  1. Then, by part (iii) of the previous proposition, we
have an exact sequence of vector spaces
o, o (d)o , 0
XEL
r(X) = k
Dk(dX) o * Dk +l ( d ) o ~ O.
Now, by Proposition 4.10, we know that Dk(d)J ~ Rk(~r ) for 0 < k < r. Once
again we have a commutative diagram with the same maps as in the proof of
Proposition 4.10. Recall that the maps in the second and third columns are
isomorphisms.
~ D k + , ( d ) ~
Dk(dX)~
, D k ( d ) ~
s 0
' Rk+,(,~r
Rk(dX)
~k ' Rk(d).
Now we see that the m a p nk is onto. Hence d is kformal. []
0
0
~ ( ~
X~L
r(X) = k
,
( ~
X~L
rlX) = k
This theorem gives additional conditions on L for free arrangements.
Suppose ~r is free. Then, by Theorem 2.10, Jzrr is free and hence kformal
for each Y e L , 0 < k < r(IO  1. Consider Y e L with r(IO > 2. By Remark 3.7,
we have
Z
X<Y
r(X) = 2
( l ~ x l  2) _> I ~ r l  r(Y).
F r o m the exact sequence we see that @
0 ~ Ra(~r) ~
F2(~x) ~ F ( ~ r ) ~ 0,
dim R3(Mr) = r(Y)  I~rl +
( l ~ x l  2).
X.<Y
r(X) = 2
Thus, for example, since ~r is 3formal, the surjectivity of
Y~L
r(r) = 3
yields the following:
Corollary 4.16. I f ~ is free and r(~r > 4, then
YcL
r(r) = 3
(3[dr[+ ~ ([~r
X~Y
r(X) = 2
~ (1~r
X~L
r(X) = 2
There are similar inequalities corresponding to kformal for k > 4.
5. AnExample
We now distinguish kformal arrangements from formal arrangements.
Example 5.1. The real 4arrangement ~/ described below is formal but not
3formal.
Free Arrangementsand Relation Spaces Let Q(~r = I~.l~ aj, where 61
~1 = X3,
OC2 ~ X 3   X4,
OC3 ~ X2,
O~4 = X 2 + X 3   2 X 4 ,
~5 = X l ,
~6 = XI + X3   2X4,
a7 = x2 + 2x3  2x,,
a8 = xl + 2x3  2x4,
69 = X1 + X 2 + X 3   2x,,
We computed L = L(~r by hand and later verified our calculations using a
computer program written by John Keaty. Note that relations a m o n g the defining
forms come from subspaces X ~ L such that r(X) < I~r Thus, in describing L,
we do not include the socalled "generic" subspaces X ~ L where r(X) = I~xl. We
use the following notation to describe the remaining elements of L. The intersection
H~ n Hj is denoted by the corresponding indices ij. We do the same for
intersections of three or more hyperplanes. Each lattice element is denoted by its maximal
index set. F o r example 14 = 17 = 147, so 147 is listed below.
R a n k 3 :
1 2 9 1 0 , 3 6 9 1 0 , 4 5 9 1 0
1 3 6 8 9 , 1 4 5 7 9 , 1 4 6 7 8 , 2 3 5 7 8 , 2 3 6 7 9 , 2 4 5 8 9 , 3 4 5 6 9
1 2 3 4 7 1 0 , 1 2 5 6 8 1 0
Rank 2:
1 2 1 0 , 1 4 7, 1 6 8 , 2 3 7 , 2 5 8 ,
3 6 9 , 4 5 9
We first show that a~r is formal. We count seven nongeneric rank 2 elements,
each an intersection of three hyperplanes. Thus we have seven relations of length
3. They are:
This leads to the matrix
~1   ~ 2   ~10 ~ O,
2a2 + a5  ~(s = O,
~t + ~ 4   ~ 7 = 0 ,
~1 + ~ 6   ~ 8
= O,
2a2 + a s   ~ 7 = O,
o~4 { o~5  o~9 = O.
This matrix has r a n k 6, and thus dim F2(.~r )  6. Recall that we have an exact
sequence
Since d is essential, dim T(~r ~ = 4, and thus dim F ( d ) = 6. Hence F2(,~ ) = F ( d )
and d is formal.
N o w show that d is not 3formal. Since ~r is formal, we have an exact sequence
0 ~ R3(.~ ) ~ (~) F2(,~/x) ~ F(,~a/) ~ 0.
XEL
r(X) = 2
N o t e that each (nongeneric) rank 2 element X e L is an intersection of three
hyperplanes, so each F2(,.~x) will have dimension 1. Thus the middle term in the
a b o v e sequence has dimension 7, and dim R 3 ( d ) = 1. In particular, R 3 ( ~ ) ~ 0.
To show that the m a p
is not surjective, we show that, for each Y e L with r(Y) = 3, we have R 3 ( d r ) = 0.
This a m o u n t s to showing that, for each such Y, the m a p O
YeL
r(Y) = 3
R3(d1") "~ R 3 ( d )
( ~
X e L (,~ly)
r(X) = 2
FE((~r)x) ~ F(dr)
is an injection.
Each T(~Cr) ~ has dimension 3, so dim F(~Cr) = I~yl  3. The chart below lists
the nongeneric rank 3 elements Y e L, along with the nongeneric rank 2 elements
of L(dv):
Rank 3 element
Contained by rank 2 element Difference 4 5 9 1 0 3 6 9 1 0
F o r each Y in the chart, the relations corresponding to the rank 2 elements of
L(dr) are linearly independent. Thus R3(~/r) = 0 for each Y of rank 3 so ~r is not
3formal. In particular, ~r is not free.
F r o m this chart, we also see that the number of nongeneric rank 2 elements of
L(~Ir) is always equal to I~r  3. Thus each ~tr is formal, so that ~t is actually
locally formal [7, Definition 2.3].
Note that we can show ~r is not free directly, by using Corollary 4.16 and the
information in the chart above. Alternatively, ~r is not free by Yuzvinsky's
inequality [6, Corollary 3.2] as well as by Terao's factorization theorem [3,
Theorem 4.137].
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4. L. Solomon and H. Terao , A formula for the characteristic polynomial of an arrangement, Adv . in Math. 64 ( 1987 ) 305  325 .
5. H. Terao , Generalized exponents of a free arrangement of hyperplanes and ShephardToddBrieskorn formula , Invent. Math . 63 ( 1981 ), 159  179 .
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